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1 year ago

What would happen to the oxygen atoms in ozone if the ozone layer were completely destroyed by ultraviolet radiation?

1 answer:

5 0

When the oxygen atoms in ozone layer were destroyed by ultraviolet radiations the individual oxygen atoms will make bond with each other and form paired oxygen molecules again to form ozone.

What is the ozone layer?

Ozone layer has a high concentration of ozone (O3) in comparison to the rest of the atmosphere, although it is still low in comparison to other gases in the stratosphere.

The ozone layer contains less than ten parts per million of ozone, whereas the normal ozone concentration in the Earth's atmosphere is around 0.3 parts per million. The ozone layer is mostly found in the lower stratosphere.

To learn more about ozone layer click the given link

brainly.com/question/1449495

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What type of bond does sodium and bromine form?

vitfil [10]

Sodium is a metal and bromine is a nonmetal so they form an ionic compound
nonmetals and nonmetals form covalent compounds

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3 years ago

8.632 nm + 8.3 nm - 30.0 nm=

Misha Larkins [42]

Answer:

13.06800 nanometers

Explanation:

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3 years ago

If 40.5 J of heat is added to a 15.4 g sample of silver, what will the change in temperature be? (Specific heat of silver is 0.2

crimeas [40]

Answer:

623.7

Explanation:

40.5*15.4=623.7

4 0

3 years ago

Criteria and constraints help determine which solutions will be applied when extracting materials from pomace in which way

DedPeter [7]

Answer:

The use of pomace for animal feed might be chosen if minimizing production costs is desired

Explanation:

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3 0

2 years ago

A 3.4 g sample of an unknown monoprotic organic acid composed of C,H, and O is burned in air to produce 8.58 grams of carbon dio

Pavlova-9 [17]

Answer:

$C_7H_6O_2$

Explanation:

Hello there!

In this case, we can divide the problem in three stages: (1) determine the empirical formula with the combustion analysis, (2) compute the molar mass of acid via the moles of the acid in the neutralization and (3) determine the molecular formula.

(1) In this case, since 8.58 g of carbon dioxide are released, we can first compute the moles of carbon in the compound:

$n_C=8.58gCO_2*\frac{1molCO_2}{44.01gCO_2}*\frac{1molC}{1molCO_2}=0.195molC$

And the moles of hydrogen due to the produced 1.50 grams of water:

$n_H=1.50gH_2O*\frac{1molH_2O}{18.02gH_2O}*\frac{2molH}{1molH_2O} =0.166molH$

Next, to compute the mass and moles of oxygen, we need to use the initial 3.4 g of the acid:

$m_O=3.4g-0.195molC*\frac{12.01gC}{1molC}-0.166molH*\frac{1.01gH}{1molH} =0.89gO\\\\n_O=0.89gO*\frac{1molO}{16.0gO}=0.0556molO$

Thus, the subscripts in the empirical formula are:

$C=\frac{0.195}{0.0556}=3.5 \\\\H=\frac{0.166}{0.0556}=3\\\\O=\frac{0.0556}{0.0556}=1\\\\C_7H_6O_2$

As they cannot be fractions.

(2) In this case, since the acid is monoprotic, we can compute the moles by multiplying the concentration and volume of KOH:

$n_{KOH}=0.279L*0.1mol/L\\\\n_{KOH}=0.0279mol$

Which are equal to the moles of the acid:

$n_{acid}=0.0279mol$

And the molar mass:

$MM_{acid}=\frac{3.4g}{0.0279mol} =121.86g/mol$

(3) Finally, since the molar mass of the empirical formula is:

7*12.01 + 6*1.01 + 2*16.00 = 122.13 g/mol

Thus, since the ratio of molar masses is 122.86/122.13 = 1, we infer that the empirical formula equals the molecular one:

$C_7H_6O_2$

Best regards!

8 0

3 years ago